# A metal has tensile strength...

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#1
A Metal has tensile strength value of 50N/m^2 and an elastic limit value of 210N/m^2.

A wire made from this metal is used to support a 10kg mass. If the cross-sectional area of the wire is 0.002m^2 :-

Calculate the force on the wire:

Multiple choice but working would be helpful!

- 160kg

- 7N

- 98.1N

- 66N
0
4 years ago
#2
(Original post by Bysteven)
A Metal has tensile strength value of 50N/m^2 and an elastic limit value of 210N/m^2.

A wire made from this metal is used to support a 10kg mass. If the cross-sectional area of the wire is 0.002m^2 :-

Calculate the force on the wire:

Multiple choice but working would be helpful!

- 160kg

- 7N

- 98.1N

- 66N
We don't give full solutions, but will help point you in the right direction.

Your units look wrong for tensile strength and elastic limit. Are you missing a 'M', or were they per ? Why is the tensile strength less than the elastic limit? Is one Young's modulus?

What is the weight of the mass?
Has the elastic limit been reached? If not, how much of the weight will be supported?
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#3
(Original post by RogerOxon)
We don't give full solutions, but will help point you in the right direction.

Your units look wrong for tensile strength and elastic limit. Are you missing a 'M', or were they per ? Why is the tensile strength less than the elastic limit? Is one Young's modulus?

What is the weight of the mass?
Has the elastic limit been reached? If not, how much of the weight will be supported?
Fairly certain that's all the information I was given, it's thrown me off too.. https://gyazo.com/55543551dd15333001d8a718ebdffdcc

Thanks for the help
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4 years ago
#4
(Original post by Bysteven)
Fairly certain that's all the information I was given, it's thrown me off too.. https://gyazo.com/55543551dd15333001d8a718ebdffdcc

Thanks for the help
Question does not make any sense with the tensile units given, but that's a distraction.

It asks for the Force acting and F=ma
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4 years ago
#5
(Original post by Bysteven)
A Metal has tensile strength value of 50N/m^2 and an elastic limit value of 210N/m^2.

A wire made from this metal is used to support a 10kg mass. If the cross-sectional area of the wire is 0.002m^2 :-

Calculate the force on the wire:

Multiple choice but working would be helpful!

- 160kg

- 7N

- 98.1N

- 66N
Assuming the wire is hanging down, and at standard gravity, the force is a very basic calculation. Perhaps you've been confused by the extraneous information provided? Why don't you present your attempt at working it out? Then we can guide you some more.
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#6
I honestly have no idea so I'm going with the option of using F = ma 10kg * 9.81, then using Stress = F/A for part B which is calculate the stress of the wire. No idea why the extraneous information is there for as it's confusing me between elastic regioning, even though the question asks no such thing.

https://gyazo.com/98e3ecaf9d976cf98f4b033f097339b2
0
4 years ago
#7
(Original post by Bysteven)
I honestly have no idea so I'm going with the option of using F = ma 10kg * 9.81, then using Stress = F/A for part B which is calculate the stress of the wire. No idea why the extraneous information is there for as it's confusing me between elastic regioning, even though the question asks no such thing.

https://gyazo.com/98e3ecaf9d976cf98f4b033f097339b2
The question is plain wrong - the units are nuts. They should be in or . I think that they meant the elastic modulus and not limit too. The elastic limit can't be above the tensile strength.

As it is, the wire breaks. The answer that I think they want is just the weight of the mass (mg).
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#8
(Original post by RogerOxon)
The question is plain wrong - the units are nuts. They should be in or . I think that they meant the elastic modulus and not limit too. The elastic limit can't be above the tensile strength.

As it is, the wire breaks. The answer that I think they want is just the weight of the mass (mg).
Also note Q3 it asks whether it's elastic or plastic and it gives True or False as a answer. I went for False i.e plastic as it was the 2nd choice and I got marked wrong. What a joke haha
0
4 years ago
#9
The question isn't wrong, nor is it nuts. It provides you with extraneous data. As stated in the OP, the question asks:

"Calculate the force on the wire".

Stresses, and cross-sectional areas are irrelevent.

F = ma = mg = 10x9.81 = 98.1 N
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4 years ago
#10
(Original post by Illgesi)
The question isn't wrong, nor is it nuts. It provides you with extraneous data. As stated in the OP, the question asks:

"Calculate the force on the wire".
Yes, it is.

The wire cannot support the weight given, so part of it will accelerate downwards with the weight.
The yield stress specified is below the elastic limit. There is no material where that is the case. They appear to be quoting Young's (/ elastic) modulus, not limit.
The claimed values are 6 orders of magnitude off the values for steel. I doubt that there's a metal with those values - they look to have been wanting values for steel, but missed by a (small) factor of one million.

(Original post by Illgesi)
Stresses, and cross-sectional areas are irrelevent.
If the wire breaks, what is the force on "the" wire?

I agree that they're simply after the weight, but, if the wire breaks, it wouldn't be that. With the values given, the wire is history.
0
4 years ago
#11
That looks like giving too much info to confuse and they just want f=ma with force as gravity.
0
4 years ago
#12
(Original post by RogerOxon)
Yes, it is.

The wire cannot support the weight given, so part of it will accelerate downwards with the weight.
The yield stress specified is below the elastic limit. There is no material where that is the case. They appear to be quoting Young's (/ elastic) modulus, not limit.
The claimed values are 6 orders of magnitude off the values for steel. I doubt that there's a metal with those values - they look to have been wanting values for steel, but missed by a (small) factor of one million.
But the issue isn't whether the values given are realistic, the question asks what the force is on the wire. It explicitly states '...is supporting...' the mass. So however instantaneously, the entire weight of the object is applied to the wire.

If the wire breaks, what is the force on "the" wire?

I agree that they're simply after the weight, but, if the wire breaks, it wouldn't be that. With the values given, the wire is history.
If the wire breaks, then the force on the wire would be zero, or rather negligible, as there would be some self-weight.
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